JAMB Mathematics Past Questions & Answers - Page 149

surface area of formula = πr\(^2\) 

If the radius is reduced then let its radius be x.

Its area is πx\(^2\) .

x : r = 2 : 5,

so \(\frac{x}{r}\) =  \(\frac{2}{5}\)

 → 5x = 2r

Hence, x = 0.4r.

Hence the area of the new circle is π (0.4r)\(^2\)  = 0.16π r\(^2\) .

The ratio of the two areas is

0.16πr\(^2\) : πr\(^2\) 

 = 0.16 : 1 = 16 : 100

= 4 : 25.

Soln. a*b = ab + a + b,

a ♦ b = a + b + 1

a*c = ac + a + c

(a*b) ♦ (a*c) = (ab + a + b + ac + a + c + 1)

= ab + ac + 2a + b + c + 1

Let the cost of living = y.

The new cost of living = \(y + \frac{15y}{100} = 1.15y\)

The food bill now = \((1 - \frac{90}{100})(1.15y)\)

= \(1.035y\)

The fractional increase in food bill = \((1.035 - 1) \times 100% = 3.5%\)

= \(\frac{35}{1000} = \frac{7}{200}\)

y = 2x² + 9x - 35

2x² + 9x = 35

x² + \(\frac{9}{2}\) = \(\frac{35}{2}\)

x² + \(\frac{9}{2}\) + \(\frac{81}{16}\) = \(\frac{35}{2}\) = \(\frac{81}{16}\)

(x + \(\frac{9}{4}\))² = \(\frac{361}{16}\)

x = \(\frac{-9}{4}\) + \(\frac{\sqrt{361}}{16}\)

x = \(\frac{-9}{4}\) + \(\frac{19}{4}\)

= 2.5 or -7

-7 < x < \(\frac{5}{2}\)

Let the present ages of father be x and son = y

five years ago, father = x - 5

son = y - 5

x - 5 = 3(y - 5)

x - 5 = 3y - 15

x - 3y = -15 + 5 = -10 ......(i)

x + y = 110......(ii)

eqn(ii) - eqn(i)

4y = 120

y = 30

sub for y = 30 in eqn(i)

x -10 = 3(30)

x -10 = 90

x = 80